Question

using chebyshevs theorem, solve these problems for a distribution with a mean of 74 and a standard deviation of 18. round k to at least 2 decimal places and final answers to at least one decimal place if needed. part 1 of 2 at least % of the values will fall between 2 and 146. part 2 of 2 at least % of the values will fall between 7 and 141.

Explanation:

Step1: Recall Chebyshev's theorem formula

The formula is $P(|X - \mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}$, where $\mu$ is the mean, $\sigma$ is the standard - deviation, and $k$ is the number of standard - deviations from the mean. First, find $k$ for each part.

Step2: Calculate $k$ for part 1

Given $\mu = 74$ and $\sigma=18$. For the interval from $2$ to $146$, we find $k$ using the formula $k=\frac{\mu - x_1}{\sigma}$ (since the interval is symmetric about the mean). Let $x_1 = 2$, then $k=\frac{74 - 2}{18}=\frac{72}{18}=4$.

Step3: Calculate the percentage for part 1

Using Chebyshev's theorem $P(|X - \mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}$. Substitute $k = 4$ into the formula: $1-\frac{1}{4^{2}}=1-\frac{1}{16}=\frac{15}{16}=0.9375 = 93.8\%$ (rounded to one decimal place).

Step4: Calculate $k$ for part 2

Given $\mu = 74$ and $\sigma = 18$. For the interval from $7$ to $141$, $k=\frac{74 - 7}{18}=\frac{67}{18}\approx3.72$.

Step5: Calculate the percentage for part 2

Using Chebyshev's theorem $P(|X - \mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}$. Substitute $k\approx3.72$ into the formula: $1-\frac{1}{3.72^{2}}=1-\frac{1}{13.8384}\approx1 - 0.0723=0.9277\approx92.8\%$ (rounded to one decimal place).

Answer:

Part 1 of 2: 93.8%
Part 2 of 2: 92.8%